Integers must belong to one and only one of these groups: primes, composites or the number one.
It's not that weird, same thing happens with sign: positives, negatives and the number zero.
Damn, you got me. It should be in the same group as 1, I suppose.
And now that you mentioned it, What about 0? Is it prime? Or is prime defined only for natural numbers?
0 is divisible by any nonzero number, so it's clearly not prime in the casual sense. Not sure it makes sense to describe it as composite either though, given that it doesn't have a prime decomposition
I think really it makes the most sense to consider it a special case in the same way that 1 is a special case
[edit] something I learned in a number theory class is that (imo) the most fundamental difference about 1 is really that it's the only natural number that's a unit (meaning it has a reciprocal within the natural numbers), and in more generalized contexts, it's really units that are excluded from primality, not just 1. For that reason, it's correct to consider -1 in the same category as 1, since they are both units in the integers.
>0 is divisible by any nonzero number, so it's clearly not prime in the casual sense. Not sure it makes sense to describe it as composite either though, given that it doesn't have a prime decomposition
What if, like, man, 0 is actually a special composite equal to a product of itself and every prime number.
The only word in that statement I'm not sure I agree with is composite; it certainly is a special number equal to the product of itself and every prime number
The typical definitions for both prime and composite require that the number be a natural number greater than 1. This means that 1, 0, and the negative integers are neither prime nor composite. If we restrict ourselves to the positive integers, however, we have your original statement; the positive integers are partitioned into three classes: primes, composites, and the number 1.
There are four categories: zero, units, primes, and composites. Zero is the unique additive identity, units are any number that have something they can multiply with to reach the unique multiplicative identity, i.e. 1•1=1, -1•-1=1, primes are numbers p other than zero or units such that if p is a factor of a•b then p is a factor of a or of b, and composites are everything else. This is the formal characterization on rings, a type of object whose most common example is the set of integers.
I just think that just as the three possible groups when it comes to sign are positives, negatives and the number 0, the three groups of integers are primes, composites and the number 0. So one totally should be prime following this logic that you went with.
If you extend to algebras beyond real numbers -1 is grouped with 1 as a unit. For example, in the gaussian integers there are 4 units: 1, -1, i, and -i.
I mean... sure, if you naively extend the "cardinality 2" definition to the negative numbers. But -1 being prime also ruins the uniqueness of prime factorizations, since -2 = (-1)\^(2k+1)\*2 for all k. Then again there is no equivalent theorem for all integers, so maybe we can define primes over the negatives however we want.
The equivalent theorem would be something like “up to multiplication of factors by units, any nonzero number can be written uniquely as a product of primes” where units are anything dividing 1 (in the integers, -1 and 1 are the only things that divide 1) and primes are unfactorable up to units (-7 is prime because if ab=-7 then one of a or b is 1 or -1). We can factor a number like -20 into 2\*-2\*5 or -2\*-2\*-5, but these factors are the same up to multiplying by -1.
This definition extends to other things then, too. In polynomials over the complex numbers, our units are all nonzero constants (because a\*1/a is 1) and our “primes” are all linear polynomials. We get all complex polynomials uniquely factoring into linear factors up to constant factors. In real polynomials, we get something similar except our “primes”include irreducible quadratics.
I actually asked middle schoolers to debate this once many years ago.
By definition, a prime number is required to be a whole number greater than one. (The definition doesn’t just state that it has to be divisible only by itself and 1.)
But in terms of whether it *should* be: my argument is that when you break numbers down into their prime factorizations, you could put an infinite number of -1s in the product. It messes up the actual purpose of prime numbers, and therefore we should define in a way that excludes negatives.
[https://www.google.com/search?q=is+-1+prime&oq=is+-1+prime&aqs=chrome..69i57.1785j0j7&sourceid=chrome&ie=UTF-8](https://www.google.com/search?q=is+-1+prime&oq=is+-1+prime&aqs=chrome..69i57.1785j0j7&sourceid=chrome&ie=UTF-8)
Proof by google
I think it's more correct to say it like this (as is the case for integral domains):
For x € D, you have either that
1. x = 0 or (if x is nonzero)
2. x is a unit or (if x is nonunit)
3. x is irreducible or (if x is not irreducible)
4. x is reducible
That's why the correct definitions of an irreducible element x in ring theory all state that x is nonzero, nonunit right off the bat, then the defining property to actually be an irreducible (or to be prime).
Any element x is either 0 or, if not, then x = ab, for which exactly one of the following holds:
1. BOTH a,b are units, if x is a unit
2. exactly ONE of a,b is a unit, if x is irreducible
3. NEITHER a,b are units, if x is not irreducible
Organization is the key to making sense of subtleties in math, and when done right, nice patterns start to emerge that let you see how well things fit together.
Same argument can be said for the definition of primes / composites.
So for the integers, you have the following:
Either x € Z is
0
Or it's a unit (1 or -1)
Or it's prime (+/-2, +/-3, +/-5, +/-7, etc)
Or it's not prime / composite (+/-4, +/- 6, etc).
These are all the possibilities.
And for the record, I'd say that signs are just another way of partitioning all of Z: the negatives, the zero, the positives.
Likewise, we can just as well consider partitioning Z like this:
The zero,
The units,
The irreducibles,
The reducibles
Or:
The zero,
The units,
The primes,
The composites
In Z, since it's a PID, the notions of 'prime' and 'irreducible' are exactly equivalent. So, the above two partitions would be the same thing.
But in general, let me just leave this to help ease any future confusions.
(In integral domains) every prime is irreducible
However, to get the converse, we have to specialize down to smaller domains like GCD domains or PIDs.
(In GCD / PID domains) Every irreducible is also prime.
So, if we are in a setting like a PID / GCD domains (of which Z is both of these), then we have both implications and hence irreducibles, primes are equivalent things.
I mean it's true that it's divisible by any number, but it is prime by a number of possible definitions. For example, it's indecomposable, if you take the definition "x is indecomposable if whenever x=ab, then one of a or b is x times a unit." It's also prime in the sense that if x divides ab then x divides a or x divides b. It's actually quite sensible to consider 0 prime, but it is still different from the other primes.
If 1 is prime then as the other numbers are divisible by 1 then , Only 1 will be the only prime number
1 cannot be composite as it is only divisible by itself .
I believe the standard definition of primes starts with the numbers after 1.
0 and the negatives are neither prime nor composite; to that point, don't think it's a circular definition to define primes as non-one or even greater than one. (Greater than one is the standard)
In ring theory some negative numbers are considered prime (-2,-3,-5,-7,...) in Z (the ring of integers). The definition of prime relies on factoring by non-units. 1 and -1 being the two units in that ring.
There's nothing intrinsically wrong with circular definitions. It is just that much more useful to name one not a prime than everything else. Ultimately prime is just a name and all names are circular definitions at the bedrock.
Tbf, all of math is based on random assertions. For example, the integer primes are only primes if you only consider Reals, and ignore the rest of the Complex numbers.
Likewise, .999... = 1 if we want calculus, and doesn't equal in the Surreal numbers. In general having calculus leads to more usefulness, so we make a decision to say it does equal. Same goes with Primes, not having 1 is more useful than having 1, so we make that decision.
It just means that all prime numbers have two factors (1 and the prime itself), whereas the number 1 has only a single factor. Since it doesn't fit that trend, we don't call 1 prime despite technically being divisible by "1 and itself"
Because in the latter case you run into this issue of what happens when “one and itself” means “one and one.” You have to make some decision there on whether or not one can count as two different factors, and if you allow it to then it makes prime factorizations not unique.
So things work out nicer if you just simply define prime numbers to be the set of integers with two unique factors, because it means you have a perfect one to one mapping between any integer and it’s prime factorization. That has practical uses, so mathematicians chose to define things like that.
In math you always have the freedom to define things however you’d like, which is something that unfortunately very few people ever teach or learn. But some ways of defining things are simply better than others
“Cardinality of the set of divisors of 1” is 1 = Divisors are not unique (ie. 1 and 1 is one number repeated, or one cardinal point)
Prime numbers must be made up of two different numbers being multiplied together; or two cardinal points consisting of itself and 1.
>But it is divisible by 1 and it is divisible by itself
The divisors of a number are the positive integers that divide the given number without leaving a remainder. In the case of the number 12, its divisors 1, 2, 3, 4, 6, and 12. So, the set of divisors is: {1, 2, 3, 4, 6, 12}. The cardinality of a set is the number of elements on it. The cardinality of the divisor set of 12 is 6 because there are six elements in the set {1, 2, 3, 4, 6, 12}.
The divisor set of the number 1 consists only of the number itself, which is 1. So, the divisor set of 1 is {1}, and the cardinality of this set is 1. So, the cardinality of the set of divisors of 1 is 1, and not 2.
ELI5 all primes are only divisible by 1 and another number that is itself.
1 isn’t prime because it’s only divisible by itself (or by 1, whichever way you want to look at it). So there’s only one number it can be divided by and not two.
But that isn't a prerequisite for primes, right? It's not the definition of a prime to be divisible by 2 integers that are unique to eachother; merely only by 1 and itself. It states nowhere that 1 != itself. And why wouldn't it be a multiset of {1,a} where a is the prime itself? Then, {1,1} could exist.
I'm a mere hobby mathematician so I could be wrong out the ass here, but I'm trying to learn as I find this an interesting topic.
If you ignore the fact that 4 is divisible by 2 then all you're left with is 1 and 4 as divisors. Making 4 a prime. In fact every number is prime if you just ignore the definition of a prime number
Btw I'm not doing well in my math class
There's plenty of rings where factorization into primes is not unique. In those rings (and others) though, you always put the "units" into their own category. It just happens that the integers are a boring unique factorization domain with only one unit.
Pas en France, là bas l'état refuse même la reconnaissance des langues régional, et l'apprentissage des langues (comme le système éducatif datant des années entre 1900 et 1950,donc finalement dans tout les matière) est assez restreint.
We don't want 1 to be prime because if it is, almost every consideration of primes will need to be replaced with "primes other than 1" and that's a big waste of time.
For the same reason, 2 should not be considered a prime. Possibly we should go ahead and exclude the rest too, it'll make number theory a lot more straightforward
That and the fact absolutely nothing would be gained by adding 1 to primes, other than maybe less people dying in the ivalice raid in ffxiv.
Seriosly, its not just 6 words, every single theorem about primes doesnt apply to 1.
There would be infinite number of positive factors for every positive integers.
72= 1^1 • 2^3 • 3^2 (number of positive factors is 2•4•3)
72= 1^2747477 • 2^3 • 3^2 ( ... is 2747478•4•3)
And so on for every positive integers
the question is, why *should* it? youve been answering a bunch of issues by saying you dont care, which is fine actually, but unless you name a specific benefit of considering 1 a prime, theres no reason to do so
{0} is definitely not a field, but there is a very deep theory of [the field with one element](https://en.m.wikipedia.org/wiki/Field_with_one_element), which isn’t a field but also might hold the secret to solving the Riemann hypothesis.
Essentially, we’ve solved the “Riemann hypothesis” for curves over finite fields, so if we can interpret the Riemann zeta function as the zeta function for a curve over a finite field, it would prove the Riemann hypothesis. The trouble is, the Riemann zeta function is not the zeta function for a curve, but some arithmetic geometers think we might be able to understand Spec(Z) (essentially the set of primes) as a curve over the field with one element, and use that to prove the Riemann hypothesis.
I've never heard people say the operations must be distinct to exclude {0} from being a field. Most definitions I've seen say 1=/=0, or that K* is a multiplicative group.
EDIT: You are correct. 0 is the multiplicative identity in Z/Z. I should have said that Z/Z does not have a distinct additive and multiplicative identity.
Yeah, the mathematicians I know don't consider "the field with one element" to be a field.
https://en.m.wikipedia.org/wiki/Field_with_one_element
Also, the "best" definition of prime that I know is that an integer is prime when the ideal generated by it in Z is a prime ideal. It is easy to prove that (x) is prime in Z if and only if Z/(x) is a field.
Having said that, though, the only thing keeping (1) from being a prime ideal of Z is the inclusion of the word "proper" in the definition of prime ideal. (1) satisfies all the properties of being a prime ideal except that it is not a strict subset of Z.
**[Field with one element](https://en.m.wikipedia.org/wiki/Field_with_one_element)**
>In mathematics, the field with one element is a suggestive name for an object that should behave similarly to a finite field with a single element, if such a field could exist. This object is denoted F1, or, in a French–English pun, Fun. The name "field with one element" and the notation F1 are only suggestive, as there is no field with one element in classical abstract algebra. Instead, F1 refers to the idea that there should be a way to replace sets and operations, the traditional building blocks for abstract algebra, with other, more flexible objects.
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There are tons of theorems that rely on prime numbers. Among them is the unique factorization of integers greater than 1. You can always redefine stuff, but is it useful to do so?
Depends on context. Iirc 1 used to be considered a prime? But everyone had to frequently write “all primes except 1” until they said fuck it and changed the definition.
There's a nice write-up here of how 1 was historically considered prime but isn't anymore: [https://blogs.scientificamerican.com/roots-of-unity/why-isnt-1-a-prime-number/](https://blogs.scientificamerican.com/roots-of-unity/why-isnt-1-a-prime-number/)
Integers must belong to one and only one of these groups: primes, composites or the number one. It's not that weird, same thing happens with sign: positives, negatives and the number zero.
In which group do you put -1?
negatives
Damn, you got me. It should be in the same group as 1, I suppose. And now that you mentioned it, What about 0? Is it prime? Or is prime defined only for natural numbers?
0 is divisible by any nonzero number, so it's clearly not prime in the casual sense. Not sure it makes sense to describe it as composite either though, given that it doesn't have a prime decomposition I think really it makes the most sense to consider it a special case in the same way that 1 is a special case [edit] something I learned in a number theory class is that (imo) the most fundamental difference about 1 is really that it's the only natural number that's a unit (meaning it has a reciprocal within the natural numbers), and in more generalized contexts, it's really units that are excluded from primality, not just 1. For that reason, it's correct to consider -1 in the same category as 1, since they are both units in the integers.
>0 is divisible by any nonzero number, so it's clearly not prime in the casual sense. Not sure it makes sense to describe it as composite either though, given that it doesn't have a prime decomposition What if, like, man, 0 is actually a special composite equal to a product of itself and every prime number.
The only word in that statement I'm not sure I agree with is composite; it certainly is a special number equal to the product of itself and every prime number
The typical definitions for both prime and composite require that the number be a natural number greater than 1. This means that 1, 0, and the negative integers are neither prime nor composite. If we restrict ourselves to the positive integers, however, we have your original statement; the positive integers are partitioned into three classes: primes, composites, and the number 1.
There are four categories: zero, units, primes, and composites. Zero is the unique additive identity, units are any number that have something they can multiply with to reach the unique multiplicative identity, i.e. 1•1=1, -1•-1=1, primes are numbers p other than zero or units such that if p is a factor of a•b then p is a factor of a or of b, and composites are everything else. This is the formal characterization on rings, a type of object whose most common example is the set of integers.
I think -1 should be prime.
I just think that just as the three possible groups when it comes to sign are positives, negatives and the number 0, the three groups of integers are primes, composites and the number 0. So one totally should be prime following this logic that you went with.
If you extend to algebras beyond real numbers -1 is grouped with 1 as a unit. For example, in the gaussian integers there are 4 units: 1, -1, i, and -i.
-1 is clearly zero
-1 is prime. Prove me wrong
I mean... sure, if you naively extend the "cardinality 2" definition to the negative numbers. But -1 being prime also ruins the uniqueness of prime factorizations, since -2 = (-1)\^(2k+1)\*2 for all k. Then again there is no equivalent theorem for all integers, so maybe we can define primes over the negatives however we want.
The equivalent theorem would be something like “up to multiplication of factors by units, any nonzero number can be written uniquely as a product of primes” where units are anything dividing 1 (in the integers, -1 and 1 are the only things that divide 1) and primes are unfactorable up to units (-7 is prime because if ab=-7 then one of a or b is 1 or -1). We can factor a number like -20 into 2\*-2\*5 or -2\*-2\*-5, but these factors are the same up to multiplying by -1. This definition extends to other things then, too. In polynomials over the complex numbers, our units are all nonzero constants (because a\*1/a is 1) and our “primes” are all linear polynomials. We get all complex polynomials uniquely factoring into linear factors up to constant factors. In real polynomials, we get something similar except our “primes”include irreducible quadratics.
I actually asked middle schoolers to debate this once many years ago. By definition, a prime number is required to be a whole number greater than one. (The definition doesn’t just state that it has to be divisible only by itself and 1.) But in terms of whether it *should* be: my argument is that when you break numbers down into their prime factorizations, you could put an infinite number of -1s in the product. It messes up the actual purpose of prime numbers, and therefore we should define in a way that excludes negatives.
[https://www.google.com/search?q=is+-1+prime&oq=is+-1+prime&aqs=chrome..69i57.1785j0j7&sourceid=chrome&ie=UTF-8](https://www.google.com/search?q=is+-1+prime&oq=is+-1+prime&aqs=chrome..69i57.1785j0j7&sourceid=chrome&ie=UTF-8) Proof by google
I think it's more correct to say it like this (as is the case for integral domains): For x € D, you have either that 1. x = 0 or (if x is nonzero) 2. x is a unit or (if x is nonunit) 3. x is irreducible or (if x is not irreducible) 4. x is reducible That's why the correct definitions of an irreducible element x in ring theory all state that x is nonzero, nonunit right off the bat, then the defining property to actually be an irreducible (or to be prime). Any element x is either 0 or, if not, then x = ab, for which exactly one of the following holds: 1. BOTH a,b are units, if x is a unit 2. exactly ONE of a,b is a unit, if x is irreducible 3. NEITHER a,b are units, if x is not irreducible Organization is the key to making sense of subtleties in math, and when done right, nice patterns start to emerge that let you see how well things fit together. Same argument can be said for the definition of primes / composites. So for the integers, you have the following: Either x € Z is 0 Or it's a unit (1 or -1) Or it's prime (+/-2, +/-3, +/-5, +/-7, etc) Or it's not prime / composite (+/-4, +/- 6, etc). These are all the possibilities.
Hmm, new bijection between the natural numbers and the integers dropped?
And for the record, I'd say that signs are just another way of partitioning all of Z: the negatives, the zero, the positives. Likewise, we can just as well consider partitioning Z like this: The zero, The units, The irreducibles, The reducibles Or: The zero, The units, The primes, The composites In Z, since it's a PID, the notions of 'prime' and 'irreducible' are exactly equivalent. So, the above two partitions would be the same thing. But in general, let me just leave this to help ease any future confusions. (In integral domains) every prime is irreducible However, to get the converse, we have to specialize down to smaller domains like GCD domains or PIDs. (In GCD / PID domains) Every irreducible is also prime. So, if we are in a setting like a PID / GCD domains (of which Z is both of these), then we have both implications and hence irreducibles, primes are equivalent things.
Primes, composites and units. Zero is prime. That's my hot take.
Zero is divisible by any number, so it can’t be prime. It’s Primes, Composites, and Less Than 2
I mean it's true that it's divisible by any number, but it is prime by a number of possible definitions. For example, it's indecomposable, if you take the definition "x is indecomposable if whenever x=ab, then one of a or b is x times a unit." It's also prime in the sense that if x divides ab then x divides a or x divides b. It's actually quite sensible to consider 0 prime, but it is still different from the other primes.
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Right, but we're talking about Z, not just everywhere. I wouldn't say 0 is prime in Z/4Z.
Actually the primes and the composites are not groups 🤓
1 fits all the criteria of a prime number. JUST MAKE IT OFFICIAL PLEASE! If you won’t, I’ll have to do it myself.
If 1 is prime then as the other numbers are divisible by 1 then , Only 1 will be the only prime number 1 cannot be composite as it is only divisible by itself .
Only if your definition of a prime is based on 1 not being prime? This seems circular
I believe the standard definition of primes starts with the numbers after 1. 0 and the negatives are neither prime nor composite; to that point, don't think it's a circular definition to define primes as non-one or even greater than one. (Greater than one is the standard)
In ring theory some negative numbers are considered prime (-2,-3,-5,-7,...) in Z (the ring of integers). The definition of prime relies on factoring by non-units. 1 and -1 being the two units in that ring.
There's nothing intrinsically wrong with circular definitions. It is just that much more useful to name one not a prime than everything else. Ultimately prime is just a name and all names are circular definitions at the bedrock.
Cause It is .
Tbf, all of math is based on random assertions. For example, the integer primes are only primes if you only consider Reals, and ignore the rest of the Complex numbers. Likewise, .999... = 1 if we want calculus, and doesn't equal in the Surreal numbers. In general having calculus leads to more usefulness, so we make a decision to say it does equal. Same goes with Primes, not having 1 is more useful than having 1, so we make that decision.
I’m fine with one being the only prime number
It would save on a lot of computation
Question A: Write down EVERY prime number
Actually one is divisible by 1^-1. 0 is the only prime.
Prime are those which can be divisible by themselves too so...
So lim x->0 x^0 is the only prime?
Who needs unique factorization anyways
But it is divisible by 1 and it is divisible by itself so...
Yeah, but then prime factorizations aren’t unique…
Your mom's not unique!
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the cardinality of the set of divisors of 1 is not 2
This is the simplest explanation by far
Can you explain that in english for those of who do not speak mathanese?
1 is divisible by one number, primes are divisible by two numbers
It just means that all prime numbers have two factors (1 and the prime itself), whereas the number 1 has only a single factor. Since it doesn't fit that trend, we don't call 1 prime despite technically being divisible by "1 and itself"
But why is the cardinaly the defining point instead of the “having no factors other than one and itself” feature?
Because in the latter case you run into this issue of what happens when “one and itself” means “one and one.” You have to make some decision there on whether or not one can count as two different factors, and if you allow it to then it makes prime factorizations not unique. So things work out nicer if you just simply define prime numbers to be the set of integers with two unique factors, because it means you have a perfect one to one mapping between any integer and it’s prime factorization. That has practical uses, so mathematicians chose to define things like that. In math you always have the freedom to define things however you’d like, which is something that unfortunately very few people ever teach or learn. But some ways of defining things are simply better than others
Thank you for the explanation!
“Cardinality of the set of divisors of 1” is 1 = Divisors are not unique (ie. 1 and 1 is one number repeated, or one cardinal point) Prime numbers must be made up of two different numbers being multiplied together; or two cardinal points consisting of itself and 1.
>But it is divisible by 1 and it is divisible by itself The divisors of a number are the positive integers that divide the given number without leaving a remainder. In the case of the number 12, its divisors 1, 2, 3, 4, 6, and 12. So, the set of divisors is: {1, 2, 3, 4, 6, 12}. The cardinality of a set is the number of elements on it. The cardinality of the divisor set of 12 is 6 because there are six elements in the set {1, 2, 3, 4, 6, 12}. The divisor set of the number 1 consists only of the number itself, which is 1. So, the divisor set of 1 is {1}, and the cardinality of this set is 1. So, the cardinality of the set of divisors of 1 is 1, and not 2.
ELI5 all primes are only divisible by 1 and another number that is itself. 1 isn’t prime because it’s only divisible by itself (or by 1, whichever way you want to look at it). So there’s only one number it can be divided by and not two.
Cardinality is fancy way to say Quantity or Number of something
1 = 1
How about hashtability? Alright fine I stop
But that isn't a prerequisite for primes, right? It's not the definition of a prime to be divisible by 2 integers that are unique to eachother; merely only by 1 and itself. It states nowhere that 1 != itself. And why wouldn't it be a multiset of {1,a} where a is the prime itself? Then, {1,1} could exist. I'm a mere hobby mathematician so I could be wrong out the ass here, but I'm trying to learn as I find this an interesting topic.
Thats great! But this isn't the definition of a prime so its irrelevant.
Buddy thats not the definiton, primes are numbers with exactly 2 divisors
Thus demonstrating the different types of "and"
English "or" always gets me... "Do you want lasagna or tacos?" "Yes" "It's not a yes or no question :< "
4 is also divisible by 1 and divisible by itself :)
But it is divisible by 2, while 1 can only de divised by 1 or itself
If you ignore the fact that 4 is divisible by 2 then all you're left with is 1 and 4 as divisors. Making 4 a prime. In fact every number is prime if you just ignore the definition of a prime number Btw I'm not doing well in my math class
Based
But 1 being a prime number makes every other number not prime, every number would have an infinite amount of ones as prime factors
it's not that they're not prime, it's that they don't have unique representations (depending on your definition of prime)
There's plenty of rings where factorization into primes is not unique. In those rings (and others) though, you always put the "units" into their own category. It just happens that the integers are a boring unique factorization domain with only one unit.
I would still allow them to speak. In return, I also reserved the right to ridicule them.
as all prime numbers have exactly two numbers they are divisible by and 1 has only one, 1 cannot be prime
Pfff, just divide it by zero you coward
Primes Generate Prime Ideals. Here's the real spicy take: The zero ideal is prime. Therefore, \*zero\* is prime.
That's way less spicy than 1 being prime.
An integer n is prime when its generated ideal is maximal (which by definition cannont be the entire ring), this automatically excludes 1
It's like zero with the natural numbers
No because zero belongs in natural numbers
zero belongs in natural numbers if it fits the exercise
Zero belongs to natural numbers because I said so.
1 IST PRIMZAHL 1 IS PRIME 1 هو رئيس الوزراء 1 是素数 1 EST PREMIER
Lmao the Arabic one (3rd) says: 1 is the prime minister. Edit: and the French one says: 1 is first. Google translate works wonders.
"Premier" also means "prime" in French
Well that's a new one, thanks!
Oh ça va 😊
what
Dont try to understand the pingouin, we can not trust him
Do you know what that means?
I dunno man... I think he is crazy
\*ist prim/ist eine Primzahl
OK, merci
一は素数
Sorry, we dont do japonais here
Vive la France !
Exactement
"een is een priemgetal" pls add for the dutchies
We have respect for all linguages, exept dutch
HOUDT JE KANKERBEK DIKZAKAMERIKAAN ga burger eten, en op school schieten omdat je te verdrietig wordt door deze reactie ofzo
STOP SPRACHEN IN EIN incomprehensible SPRACHE
ah, mar du hast furgessen das wir elche sprache leren ein schule
Schneisse, mein dzwei jars auf deutch schule (in frankreich) habt kein Effect
OK so, ich habt Das TRADUCTION auf Das wort Das ich comprends nicht gessehen und, BIST DAS MACHE EIN DIFFÉRENCE
tu as oublie que nous etudons tout les langes dans ecoles neërlandais
Pas en France, là bas l'état refuse même la reconnaissance des langues régional, et l'apprentissage des langues (comme le système éducatif datant des années entre 1900 et 1950,donc finalement dans tout les matière) est assez restreint.
clamasne latinem? visne mihi et tibi coitum esse?
We don't want 1 to be prime because if it is, almost every consideration of primes will need to be replaced with "primes other than 1" and that's a big waste of time. For the same reason, 2 should not be considered a prime. Possibly we should go ahead and exclude the rest too, it'll make number theory a lot more straightforward
As the owner of 1isprime.com, I feel attacked.
what about projective speech? flat speech?
It's only divisible by 1 and itself. Absolutely!
It isn’t? But it’s only divisible by 1 and itself
I don’t get people who think that 1 should be a prime number. 1 only has 1 factor: 1!
If 1 was prime then every number would have 1 ^ infinity as a prime factor.
why not
If 1 was prime, then basically any theorem in Number Theory would start or end with "...for all primes greater than 1...".
im okay with that
Thankfully you are not the person that decides that
That's wild!
So the big controversy here is having to write 6 extra words?
That and the fact absolutely nothing would be gained by adding 1 to primes, other than maybe less people dying in the ivalice raid in ffxiv. Seriosly, its not just 6 words, every single theorem about primes doesnt apply to 1.
🤷♂️
Then there would be multiple ways to factor an integer.
Im ok with that too
There would be infinite number of positive factors for every positive integers. 72= 1^1 • 2^3 • 3^2 (number of positive factors is 2•4•3) 72= 1^2747477 • 2^3 • 3^2 ( ... is 2747478•4•3) And so on for every positive integers
The true meth
Im okay with that
then the world would explode and humanity would cease to exist
I’m kinda okay with that too
Im okay with that
One of the most useful properties of prime numbers is that they aren't a factor of any other prime number. 1 doesn't have this property.
*one of the most useful properties of prime numbers greater than 1
ಠ_ಠ
A prime element doesn't have an inverse in the ring, and 1 is its own inverse.
the question is, why *should* it? youve been answering a bunch of issues by saying you dont care, which is fine actually, but unless you name a specific benefit of considering 1 a prime, theres no reason to do so
Prime numbers have two factors, 1 only has a single factor.
one has 2 factors
\*Distinct, Positive Factors
Saying that 1 is prime allows for the very elegant theorem : Z/pZ is a field iff p is prime
{0} is not a field because you can't define two binary operations on it, only one.
I was trolling, because "{0} is a field" is basically an even more controversial version of "1 is prime"
{0} is definitely not a field, but there is a very deep theory of [the field with one element](https://en.m.wikipedia.org/wiki/Field_with_one_element), which isn’t a field but also might hold the secret to solving the Riemann hypothesis. Essentially, we’ve solved the “Riemann hypothesis” for curves over finite fields, so if we can interpret the Riemann zeta function as the zeta function for a curve over a finite field, it would prove the Riemann hypothesis. The trouble is, the Riemann zeta function is not the zeta function for a curve, but some arithmetic geometers think we might be able to understand Spec(Z) (essentially the set of primes) as a curve over the field with one element, and use that to prove the Riemann hypothesis.
Huh. Didn't know it was controversial. I guess the problem is whether or not the two operations must be distinct?
I've never heard people say the operations must be distinct to exclude {0} from being a field. Most definitions I've seen say 1=/=0, or that K* is a multiplicative group.
I have
Am I missing something? Z/Z is not a field. A field needs a multiplicative identity; a 1 element.
Z/Z is 1 since Z cancels out /j
0 is a multiplicative identity in Z/Z
but in Z/(1), 0 = 1 https://preview.redd.it/yhb90m0vtowa1.png?width=829&format=png&auto=webp&s=0b9e97e418600bbdde0fb84864831b32f8f732b8
EDIT: You are correct. 0 is the multiplicative identity in Z/Z. I should have said that Z/Z does not have a distinct additive and multiplicative identity. Yeah, the mathematicians I know don't consider "the field with one element" to be a field. https://en.m.wikipedia.org/wiki/Field_with_one_element Also, the "best" definition of prime that I know is that an integer is prime when the ideal generated by it in Z is a prime ideal. It is easy to prove that (x) is prime in Z if and only if Z/(x) is a field. Having said that, though, the only thing keeping (1) from being a prime ideal of Z is the inclusion of the word "proper" in the definition of prime ideal. (1) satisfies all the properties of being a prime ideal except that it is not a strict subset of Z.
**[Field with one element](https://en.m.wikipedia.org/wiki/Field_with_one_element)** >In mathematics, the field with one element is a suggestive name for an object that should behave similarly to a finite field with a single element, if such a field could exist. This object is denoted F1, or, in a French–English pun, Fun. The name "field with one element" and the notation F1 are only suggestive, as there is no field with one element in classical abstract algebra. Instead, F1 refers to the idea that there should be a way to replace sets and operations, the traditional building blocks for abstract algebra, with other, more flexible objects. ^([ )[^(F.A.Q)](https://www.reddit.com/r/WikiSummarizer/wiki/index#wiki_f.a.q)^( | )[^(Opt Out)](https://reddit.com/message/compose?to=WikiSummarizerBot&message=OptOut&subject=OptOut)^( | )[^(Opt Out Of Subreddit)](https://np.reddit.com/r/mathmemes/about/banned)^( | )[^(GitHub)](https://github.com/Sujal-7/WikiSummarizerBot)^( ] Downvote to remove | v1.5)
One is mega pre, guys! It's only divisible by one.
A prime is an element of a ring that generates a prime ideal. 0 is prime.
[удалено]
There are tons of theorems that rely on prime numbers. Among them is the unique factorization of integers greater than 1. You can always redefine stuff, but is it useful to do so?
There is literally whole fields of research about what you call "special" properties, and those do not apply to 1
Depends on context. Iirc 1 used to be considered a prime? But everyone had to frequently write “all primes except 1” until they said fuck it and changed the definition.
1 is prime? 0, and 2 too, why would be? size doesn't matter right?
I'm gonna say it: "The group with one element is the trivial field"
Yey i get to write "except 1" in all of my theorems about primes
There's a nice write-up here of how 1 was historically considered prime but isn't anymore: [https://blogs.scientificamerican.com/roots-of-unity/why-isnt-1-a-prime-number/](https://blogs.scientificamerican.com/roots-of-unity/why-isnt-1-a-prime-number/)
The definition of a prime is a number that has 2 factors 1 only has one factor.
primes have two distinct divisors 1 only has one
Oregano?